Solve 2+2i/2+i*(2-1)/(2-1)

$$\frac { 2 + 2 i } { 2 + i } \cdot \frac { ( 2 - 1 ) } { ( 2 - 1 ) }$$

Evaluate

$$\frac{6}{5}+\frac{2}{5}i=1.2+0.4i$$

Solution Steps

Divide 2-1 by 2-1 to get 1.

$$\frac{2+2i}{2+i}\times 1$$

Multiply both numerator and denominator of \frac{2+2i}{2+i} by the complex conjugate of the denominator, 2-i.

$$\frac{\left(2+2i\right)\left(2-i\right)}{\left(2+i\right)\left(2-i\right)}\times 1$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2+2i\right)\left(2-i\right)}{2^{2}-i^{2}}\times 1$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{\left(2+2i\right)\left(2-i\right)}{5}\times 1$$

Multiply complex numbers 2+2i and 2-i like you multiply binomials.

$$\frac{2\times 2+2\left(-i\right)+2i\times 2+2\left(-1\right)i^{2}}{5}\times 1$$

By definition, i^{2} is -1.

$$\frac{2\times 2+2\left(-i\right)+2i\times 2+2\left(-1\right)\left(-1\right)}{5}\times 1$$

Do the multiplications in 2\times 2+2\left(-i\right)+2i\times 2+2\left(-1\right)\left(-1\right).

$$\frac{4-2i+4i+2}{5}\times 1$$

Combine the real and imaginary parts in 4-2i+4i+2.

$$\frac{4+2+\left(-2+4\right)i}{5}\times 1$$

Do the additions in 4+2+\left(-2+4\right)i.

$$\frac{6+2i}{5}\times 1$$

Divide 6+2i by 5 to get \frac{6}{5}+\frac{2}{5}i.

$$\left(\frac{6}{5}+\frac{2}{5}i\right)\times 1$$

Multiply \frac{6}{5}+\frac{2}{5}i and 1 to get \frac{6}{5}+\frac{2}{5}i.

$$\frac{6}{5}+\frac{2}{5}i$$

Real Part

$$\frac{6}{5} = 1\frac{1}{5} = 1.2$$

Solution Steps

Divide 2-1 by 2-1 to get 1.

$$Re(\frac{2+2i}{2+i}\times 1)$$

Multiply both numerator and denominator of \frac{2+2i}{2+i} by the complex conjugate of the denominator, 2-i.

$$Re(\frac{\left(2+2i\right)\left(2-i\right)}{\left(2+i\right)\left(2-i\right)}\times 1)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{\left(2+2i\right)\left(2-i\right)}{2^{2}-i^{2}}\times 1)$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{\left(2+2i\right)\left(2-i\right)}{5}\times 1)$$

Multiply complex numbers 2+2i and 2-i like you multiply binomials.

$$Re(\frac{2\times 2+2\left(-i\right)+2i\times 2+2\left(-1\right)i^{2}}{5}\times 1)$$

By definition, i^{2} is -1.

$$Re(\frac{2\times 2+2\left(-i\right)+2i\times 2+2\left(-1\right)\left(-1\right)}{5}\times 1)$$

Do the multiplications in 2\times 2+2\left(-i\right)+2i\times 2+2\left(-1\right)\left(-1\right).

$$Re(\frac{4-2i+4i+2}{5}\times 1)$$

Combine the real and imaginary parts in 4-2i+4i+2.

$$Re(\frac{4+2+\left(-2+4\right)i}{5}\times 1)$$

Do the additions in 4+2+\left(-2+4\right)i.

$$Re(\frac{6+2i}{5}\times 1)$$

Divide 6+2i by 5 to get \frac{6}{5}+\frac{2}{5}i.

$$Re(\left(\frac{6}{5}+\frac{2}{5}i\right)\times 1)$$

Multiply \frac{6}{5}+\frac{2}{5}i and 1 to get \frac{6}{5}+\frac{2}{5}i.

$$Re(\frac{6}{5}+\frac{2}{5}i)$$

The real part of \frac{6}{5}+\frac{2}{5}i is \frac{6}{5}.

$$\frac{6}{5}$$